Select random values from each group, SQL

Question

I have a project through which I'm creating a game powered by a database.

The database has data entered like this:

(ID, Name) || (1, PhotoID),(1,PhotoID),(1,PhotoID),(2,PhotoID),(2,PhotoID) and so on. There are thousands of entries.

This is my current SQL statement:

$sql = "SELECT TOP 8 * FROM Image WHERE Hidden = '0' ORDER BY NEWID()";

But this can also produce results with matching IDs, where I need to have each result have a unique ID (that is I need one result from each group).

How can I change my query to grab one result from each group?

Thanks!

Answer 1

Since ORDER BY NEWID() will result in tablescan anyway, you might use row_number() to isolate first in group:

; with randomizer as (
  select id,
         name,
         row_number() over (partition by id
                            order by newid()) rn
    from Image
   where hidden = 0
)
select top 8
       id,
       name
  from randomizer
 where rn = 1
-- Added by mellamokb's suggestion to allow groups to be randomized
order by newid()

Sql Fiddle playground thanks to mellamokb.

Answer 2

Looks like this may work, but I can't vouch for performance:

SELECT TOP 8 ID,
  (select top 1 name from image i2
   where i2.id = i1.id order by newid())
FROM Image i1
WHERE hidden = '0'
group by ID
ORDER BY NEWID();

Demo: http://www.sqlfiddle.com/#!3/657ad/6

Answer 3

If you have an index on the ID column and want to take advantage of the index and avoid a full table scan, do your randomization on the key values first:

WITH IDs AS
(
  SELECT DISTINCT ID
  FROM Image
  WHERE Hidden = '0'
),
SequencedIDs AS
(
  SELECT ID, ROW_NUMBER() OVER (ORDER BY NEWID()) AS Seq
  FROM IDs
),
ImageGroups AS
(
  SELECT i.*, ROW_NUMBER() OVER (PARTITION BY i.ID ORDER BY NEWID()) Seq
  FROM SequencedIDs s
  INNER JOIN Image i
    ON i.ID = s.ID
  WHERE s.Seq < 8
  AND i.Hidden = '0'
)
SELECT *
FROM ImageGroups
WHERE Seq = 1

This should drastically reduce the cost over the table scan approach, although I don't have a schema big enough that I can test with - so try running some statistics in SSMS and make sure ID is actually indexed for this to be effective.

Answer 4

select * from (select * from photos order by rand()) as _SUB group by _SUB.id;

Answer 5

 select ID, Name from (select ID, Name, row_number() over 
 (partition by ID, Name order by ID) as ranker from Image where Hidden = 0 ) Z where ranker = 1
 order by newID()